\(A=\frac{x}{4}+\frac{4}{x}=\frac{x^2+16}{4x}=\frac{x^2+17x+16-17x}{4x}=-\frac{17}{4}+\frac{\left(x+1\right)\left(x+16\right)}{4x}\)
Do \(-1\le x< 0\Rightarrow\left\{{}\begin{matrix}x+16>0\\x+1\ge0\\4x< 0\end{matrix}\right.\) \(\Rightarrow\frac{\left(x+1\right)\left(x+16\right)}{4x}\le0\)
\(\Rightarrow A\le-\frac{17}{4}\)
\(A_{max}=-\frac{17}{4}\) khi \(x=-1\)